Question

Help I am stuck!! : http://nimb.ws/faXEPi

Answer 1

laughing out loud, I like your commentary on these questions.

Answer 2

Lol thank you ^^.

Answer 3

So for 14, you have \[y^2+x^2-16y+39=0\]\[y^2-x^2-9=0\] This is going to be a little long.

Answer 4

@mathmate Mind looking over my solution? We're ready to start, haha.

Answer 5

Keep going, you're on the right track!

Answer 6

First put your two equations into the proper equation of a circle. You can do this by completing the square. \[y^2+x^2 -16y = -39\]\[(y^2-16y)+x^2=-39\]\[(y^2 -16y +64)+x^2 =-39+64\]\[(y-8)^2 +x^2 = 25\]

Answer 7

Now let's do equation 2.\[y^2-x^2-9 = 0\]\[y^2-x^2=9\]

Answer 8

Completing the square is a good start, but it's not quite that important here, I just wanted to show you that if you had a more complicated problem, and you didn't know how to expand your function, the best method would be to put it in the equation of a circle found by completing the square.

Let's try it again.

Answer 9

Thank You this explanation is really helping :).

Answer 10

\[y^2+x^2-16y=-39\]\[y^2-x^2=9\] you will notice that by adding both of these equations together, the \(x^2\) will cancel out. You will be left with:\[2y^2-16y=-30\]Solving this quadratic, we want to move \(-39\) over to the left side. \[2y^2-16y+30=0\]\[y^2-8y+15=0\]\[(y-5)(y-3)=0\]

Answer 11

Therefore you get two solutions for y: \(y=5~,~ y=3\)

Answer 12

Now plugging these back in to your second equation, you can find the values of x :)

Answer 13

Thank You so much ! You are awesome!!

Answer 14

\(\color{blue}{\Large\checkmark}{Way~to~go!}\)

Answer 15

\[y^2-x^2=9\]\[-x^2=9-y^2\]\[x^2=y^2-9~\sf ~\text{where y = 5,3}\]\[x^2=(5)^2-9 \implies x=\pm\sqrt{25-9} = \pm 4\]\[x^2=(3)^2 - 9 \implies x= \pm\sqrt{9-9}=0\]\(\therefore\) Your intersection will be?